EVT Example 1
[Lecture 10] How does EVT fail when \(f\) is not continuous?
Solution
Let’s take \(f(x) = \frac{1}{x}\) on \([0,1]\). Note here that we can’t really talk about continuity at \(x = 0\) since \(f\) is not even defined on \(x = 0\). Thus, define \(f\) as follows
$$
\begin{align}
f(x) =
\begin{cases}
\frac{1}{x}, & \text{if } 0 < x \leq 1, \\
0, & \text{if } x = 0.
\end{cases}
\end{align}
$$
We showed here that \(f\) is not continuous at \(x = 0\). Now, notice that \(f\) doesn’t attain a maximum in \([0,1]\). It is unbounded since as \(x \to 0\), \(f(x) \to \infty\). Hence, the continuity condition is essential.
EVT Example 2
[Lecture 10] How does EVT fail when \([a,b]\) is not bounded?
Solution
EVT Example 3
[Lecture 10] How does EVT fail when \([a,b]\) is not closed?
Solution
References
- Introduction to Analysis, An, 4th edition by William Wade
- Lecture Notes by Professor Chun Kit Lai